Tag: forced vibration

Questions Related to forced vibration

Find the time period of small oscillations of the following systems. 

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. A metre stick suspended through the 20 cm mark.

  2. A ring of mass m and radius r suspended through a point on its perphery.

  3. A uniform square plate of edge a suspended through a corner.

  4. A uniform disc of mass m and radius r suspended through a point r/2 away from the centre.

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Correct Option: A

A uniform circular disc of radius $R$ oscillates about a horizontal axis in its own plane. The distance of the axis from the center for the period of oscillation is maximum, will be :

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. $R$

  2. $\dfrac{R}{\sqrt 2}$

  3. $\dfrac{R}{3}$

  4. $\dfrac{R}{4}$

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Correct Option: B

Find the frequency of oscillation of the spheres

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. $\frac{1}{{2\pi }}\sqrt {\dfrac{{35K}}{{46m}}} $

  2. $\frac{1}{{2\pi }}\sqrt {\dfrac{{46K}}{{35m}}} $

  3. $\frac{1}{{2\pi }}\sqrt {\dfrac{{25K}}{{46m}}} $

  4. $\frac{1}{{2\pi }}\sqrt {\dfrac{{21K}}{{46m}}} $

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Correct Option: A

Find the frequency of oscillation of the spheres

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. $ \dfrac { 1 }{ 2\pi } \sqrt { \dfrac { 35K }{ 46m } } $

  2. $ \dfrac { 1 }{ 2\pi } \sqrt { \dfrac { 46K }{ 35m } } $

  3. $ \dfrac { 1 }{ 2\pi } \sqrt { \dfrac { 25K }{ 46m } } $

  4. $ \dfrac { 1 }{ 2\pi } \sqrt { \dfrac {21K }{ 46m } } $

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Correct Option: C

A particle of mass m is in one dimensional potential field and its potential energy is given by the following equation U(x)=${U _0}\left( {1 - \cos \;aX} \right)\;where\;{U _0}$ and $\alpha $ constants.The period of the particle for small oscillations near the equilibrium will be-

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. $2\pi \sqrt {\dfrac{{m{\alpha ^2}}}{{m{\alpha ^2}{U _0}}}} $

  2. $2\pi \sqrt {m{\alpha ^2}{U _0}} $

  3. $2\pi \sqrt {\dfrac{m}{{{\alpha ^2}{U _0}}}} $

  4. $2\pi \sqrt {\dfrac{{{\alpha ^2}{U _0}}}{m}} $

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Correct Option: A

A boy is playing on a swing in sitting position. the time period of oscillation of the swing is T, if the boy stands up, the time period of oscillation of the spring will be:

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. $T$

  2. $ Less \ than T$

  3. $More\  than T$

  4. such as cannot be predicted

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Correct Option: A

The time taken by a particle performing S.H.M. to pass from point $ A  $ to $  B  $ where its velocities are same is $2$ seconds. After another 2 seconds it returns to $ \mathrm{B}  $ . The time period of oscillation is (in seconds):

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. $2$

  2. $4$

  3. $6$

  4. $8$

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Correct Option: D
Explanation:

According to the question, a and b points are such that they are

 located at same distances from the equilibrium position. 
Here also it is said that velocity is same 
,i.e; not only the magnitude but also the directions are same.
 So, total time period of oscillation $= 2×$( time taken to go from a to b
 + the next time taken to return at b) $= 2×(2+2)= 8$ sec.

A student measures the time period of oscillation of a simple pendulum. He uses the data to estimate the acceleration due to gravity 9g) at that place. If the maximum percentage error in measurement of length pendulum and that in time are $ e _{1} $ and $ e _{2} $ respectively then percentage error estimation of ''g'' is :

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. $

    e _{1}+2 e _{2}

    $

  2. $

    2 e 1+e 2

    $

  3. $

    e 1+e _{2}

    $

  4. $

    e 1-e _{2}

    $

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Correct Option: A

The phase of particle in SHM is found to increase by $14 \pi$ in 3.5 sec. Its frequency of oscillation is

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. $2 Hz$

  2. $1/2 Hz$

  3. $1 Hz$

  4. $2 \pi Hz$

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Correct Option: B

Frequency of oscillation of a body is $6\;Hz$ when force $F _1$ is applied and $8\;Hz$ when $F _2$ is applied. If both forces $F _1\;&\;F _2$ are applied together then, the frequency of oscillation is :

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. $14\;Hz$

  2. $2\;Hz$

  3. $10\;Hz$

  4. $10\surd{2}\;Hz$

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Correct Option: C
Explanation:

According to question,

$F _1=-K _1\,x\;\&\;F _2=-K _2\,x$

So $n _1=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{K _1}{m}}=6\;Hz;$

$n _2=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{K _2}{m}}=8\;Hz$

Now $F=F _1+F _2=-(K _1+K _2)x$

Therefore $n=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{K _1+K _2}{m}}$

$\Rightarrow n=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{4\pi^2\,n^2 _1\,m+4\pi^2\,n^2 _2\,m}{m}}$$=\sqrt{n _1^2+n _2^2}=\sqrt{8^2+6^2}$

$=10\;Hz$